Do Gravitating Bodies Warp the Fabric of Space?

[Buckethead]

If you think about space time as a baloon where the stretchiness of the balloon at a spot on its surface is determined by its mass/energy density, then the surface of the balloon will be dimpled. The rate of time and the spatial dimensions are all determined by the radius of the dimple. Motion across the surface of the baloon means that you will be moving through dimples in space time as well as causing a dimple to propagate over the surface.

This is a great analogy. I knew about the radius of the baloon represented time and the surface a 3D space, but when you consider that massive objects are actually living in the past since their clocks are slowed relative to empty space, they will of course dimple the surface partially into the past. Great visual!

 

[Passionflower]

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

I understand this is from a book by L Epstein. It clearly was never professionally reviewed as the concept of curved proper time is sheer nonsense.

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling. Curvature of spacetime simply causes objects to accelerate with respect to each other without any need for proper acceleration.

 

[A.T.]

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

I understand this is from a book by L Epstein. It clearly was never professionally reviewed

Rickard Jonsson has derived the math of Epstein embeddings:
http://www.relativitet.se/Webtheses/lic.pdf (Chapter 6, page 53)

as the concept of curved proper time is sheer nonsense.

A single dimension cannot be intrinsically curved alone, that's why "curved time" is in quotes. But intrinsic curvature is not even necessary to have a "gravitational pull". The "gravitational pull" is related to the 1st derivates of the metric, while tidal forces are related to 2nd derivates (curvature). And the spacetime in the picture in fact doesn't have any intrinsic curvature. I agree that "curved time" is not a good title for the illustration.

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling.

So you think, greater proper acceleration means slower clock rate? Some counter-examples:

- A clock resting in the Earths center experiences less proper acceleration but runs slower than a clock resting on the surface.

- You can have a two clocks resting (relative to each other) in an uniform gravitational field experiencing the same proper acceleration, but running at different rates.

Curvature of spacetime simply causes objects to accelerate with respect to each other without any need for proper acceleration.

Yes.

 

[Passionflower]

- A clock resting in the Earths center experiences less proper acceleration but runs slower than a clock resting on the surface.

- You can have a two clocks resting (relative to each other) in an uniform gravitational field experiencing the same proper acceleration, but running at different rates.

A clock in the center of the Earth accelerates away from all directions because all the mass surrounding it tries to attract it. Proper acceleration does not necessarily result in relative movement.

A uniform gravitational field is a red herring, as the question often becomes how uniform is a uniform gravitational field really.

 

[A.T.]

A clock in the center of the Earth accelerates away from all directions because all the mass surrounding it tries to attract it.

That is just a very complicated way to say that the proper acceleration of a clock resting in the center is zero. This is less than the proper acceleration of a clock resting on the surface. Yet the center clock runs slower than the surface clock.

This is shows that your idea, that greater proper acceleration causes a slower clock rate, is false. Here your statement that I was objecting to:

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling.

Do yon now understand, that it isn't the difference in proper acceleration that determines gravitational time dilation between two clocks?

 

[Passionflower]

An object can be both stationary and accelerating.

 

[Dale]

Not proper acceleration, that can only have one value, the one measured by an accelerometer. An accelerometer at the center of the Earth reads 0, and accelerometer at the surface of the Earth reads 9.8 m/s² upwards. A.T.'s counterexample is correct, gravitational time dilation is not due to differences in proper acceleration the way you suggest.

The uniform field is also a good counter example. Suppose you have an ideal gravitational field where the proper acceleration of a stationary particle is everywhere constant. In such a field a light pulse going "up" would be gravitationally red-shifted and therefore there would be gravitational time dilation despite the fact that the proper acceleration is constant.

 

[A.T.]

An object can be both stationary and accelerating.

I don't quite see how this addresses my counter example to your claim that greater proper acceleration causes a slower clock rate.

 

[starthaus]

That is just a very complicated way to say that the proper acceleration of a clock resting in the center is zero. This is less than the proper acceleration of a clock resting on the surface. Yet the center clock runs slower than the surface clock.

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_2=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field $$\Phi_1-\Phi_2$$.

Generalization:

At a distance $$r<R$$ from the center of the sphere, inside the sphere, the gravitational potential is:

$$\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})$$

The above gives:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1$$

For $$r=0$$ (clock2 at the center of the Earth) you recover the results from above.

For $$r=R$$ you get the expected:

$$\frac{d\tau_1}{d\tau_2}=1$$

 

[TCS]

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-2\frac{\Phi}{c^2})(cdt)^2+(1-2\frac{\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-2\frac{\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-2\frac{\Phi_1}{c^2}}{1-2\frac{\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_1=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-1/2\frac{GM}{rc^2}<1$$

So, $$f_1>f_2$$

In addition, the time dilation depends on the gradient of the gravitational field $$\Phi_1-\Phi_2$$, i.e., it depends on acceleration.

Shouldn't phi 1 be zero. I think that you are missing a little r in your calculation.

 

[starthaus]

Shouldn't phi 1 be zero. I think that you are missing a little r in your calculation.

No, the calculation is correct.

 

[TCS]

This link shows that gravitational time dilation is proportion to small g, which should be zero at the center of the earth.

https://www.physicsforums.com/library.php?do=view_item&itemid=166

 

[starthaus]

This link shows that gravitational time dilation is proportion to small g, which should be zero at the center of the earth.

https://www.physicsforums.com/library.php?do=view_item&itemid=166

The formulas on the link you cited are valid for outside the Earth. The potentials inside the Earth are different.You need to be careful with what expressions you plug in into your calculations. The calculations I showed are correct.

 

[TCS]

The formulas on the link you cited are valid for outside the Earth. The potentials inside the Earth are different.You need to be careful with what expressions you plug in into your calculations. The calculations I showed are correct.

That does make more sense to me because I had thought that time the contraction was determined by the energy density and those equations seemd to contradict my belief.

Is the difference related to change in potential energy of the clock?

 

[Dale]

Is the difference related to change in potential energy of the clock?

Yes. E.g. as a photon goes up it gains potential energy, loses kinetic energy, and therefore becomes redshifted. This indicates that time is slower lower in the potential well.

 

[Passionflower]

Not proper acceleration, that can only have one value, the one measured by an accelerometer. An accelerometer at the center of the Earth reads 0, and accelerometer at the surface of the Earth reads 9.8 m/s² upwards. A.T.'s counterexample is correct, gravitational time dilation is not due to differences in proper acceleration the way you suggest.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

An accelerometer at the center of the Earth reads 0

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

The uniform field is also a good counter example. Suppose you have an ideal gravitational field where the proper acceleration of a stationary particle is everywhere constant. In such a field a light pulse going "up" would be gravitationally red-shifted and therefore there would be gravitational time dilation despite the fact that the proper acceleration is constant.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Yes. E.g. as a photon goes up it gains potential energy, loses kinetic energy, and therefore becomes redshifted. This indicates that time is slower lower in the potential well.

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

 

[Dale]

Passionflower, you have a strong habit of making factually wrong statements and then when you are corrected by someone who actually knows what they are talking about making it worse by persisting in your error and trying to support it with completely irrelevant statements. You did it in the thread where you were confusing path length with distance and now in this thread.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

Irrelevant. Proper acceleration is defined along a worldline, stresses are not.

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

Also irrelevant. As you yourself mention, measuring acceleration in one direction is a limitation of a specific type of accelerometer. Similarly, not operating properly at the center of the Earth is a limitation of a specific device, not a limitation of nature. In relativity when we talk about clocks, rods, or accelerometers we are always speaking of ideal devices.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Sure, no problem. A scalar gravitational potential can be defined in a static spacetime, such as the Schwarzschild solution. "Up" is the direction of the gradient of the potential, and the region around the minimum of a potential is described as a "well".

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

And how would you make an experiment that would test this distinction? As far as I can tell it is two different ways of saying the same thing.

Again, all of this is in response to your factually incorrect statement that "The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling." A.T. provided two correct counter examples and none of your posts since then have addressed anything relevant, which is the exact same behavior as in the other thread.

 

[yuiop]

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_2=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field $$\Phi_1-\Phi_2$$.

Generalization:

At a distance $$r<R$$ from the center of the sphere, inside the sphere, the gravitational potential is:

$$\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})$$

The above gives:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1$$

For $$r=0$$ (clock2 at the center of the Earth) you recover the results from above.

For $$r=R$$ you get the expected:

$$\frac{d\tau_1}{d\tau_2}=1$$

Here is a slightly better equation based on the interior Schwarzschild solution for $$dr=d\theta=d\phi=0$$ :

$$\frac{d\tau}{dt}=\frac{3}{2}\sqrt{1-\frac{2GM}{c^2R}}-\frac{1}{2}\sqrt{1-\frac{2GMr^2}{c^2R^3}}$$

where $$d\tau$$ is the proper clock rate of a clock located at r, $$dt$$ is the clock rate of a clock located very distant from the gravitational body and R is the the radius of the massive body with mass M, valid for $$r<=R$$. This equation has the slightly curious result that when $$R <(9/4)GM/c^2$$ the proper time of a clock at the centre runs backwards compared to the distant clock.

The equation assumes an even density distribution within the massive body and can be be expressed more generally for uneven density distributions (using units of G=c=1) as:

$$\frac{d\tau}{dt}=3/2\sqrt{1-2M/R}-1/2\sqrt{1-2 p (4/3)\pi r^2}$$

where p is the average density of the mass enclosed within a sphere of radius r where the clock is located. To compare the clock rates of two clocks located at a radius less than or equal to the radius of the gravitational body the following equation can be used:

$$\frac{d\tau_1}{d\tau_2}= \frac{3\sqrt{1-2M/R}-\sqrt{1-2 p_1 (4/3)\pi r_1^2}}{3\sqrt{1-2M/R}-\sqrt{1-2 p_2 (4/3)\pi r_2^2}}$$

For the case that one clock is located at the surface ($$dt_1 = dt_s$$) of the massive body ($$r_1 = R$$) and the other clock is located at the centre ($$dt_2= dt_c$$) of the massive body ($$r_2 = 0$$) the equation simplifies to:

$$\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}$$

which has a value of greater than unity for all $$2M<R<\infty$$

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

 

[starthaus]

$$\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}$$

which has a value of greater than unity for all $$2M<R<\infty$$

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:$$\frac{d\tau_s}{d\tau_c}<1$$.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

 

[Passionflower]

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

That is certainly true.

However, in order to calculate such situations one has to start with some kind of perfect fluid solution as the Earth is not hollow. Geodesic paths, due to Ricci curvature, tend to get in in each others way in the center and create havoc to say the least.

 

[Dale]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

 

[yuiop]

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:

$$\frac{d\tau_s}{d\tau_c}<1$$.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

$$\frac{d\tau_s}{d\tau_c}>1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

From this you should get:

$$\frac{f_1}{f_2} > 1$$

For example the clock rate ($$d\tau_s$$) of a clock on the surface of a body with a radius R=8M is 0.8666 seconds for every second that passes on a clock at infinity.

The clock rate ($$d\tau_c$$) of a clock at the centre of the same body is 0.7990 seconds for every second that passes on a clock at infinity (and slower than the surface clock).

The ratio is this example is:

$$\frac{d\tau_s}{d\tau_c} = \frac{0.8666}{0.7990} = 1.084 >1$$

Maybe there is a mistake in your calculations?

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

Really now, who is talking about particular solutions you or I?

Do you realize that the EM forces in all atoms inside the Earth's have to resist the tendency of gravity to collapse the Earth into a black hole. That resistance is called proper acceleration.

 

[starthaus]

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

$$\frac{d\tau_s}{d\tau_c}>1$$

The above signifies that the period of the clock at the surface is greater than the one at the ceneter of the Earth. This is incorrect.

From this you should get:

$$\frac{f_1}{f_2} > 1$$

Nope, you got this backwards:

$$\frac{d\tau_s}{d\tau_c}>1$$

implies:

$$\frac{f_1}{f_2} < 1$$

which is clearly wrong. Remember, clock frequency is the inverse of clock period.

 

[A.T.]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

If you think that the proper acceleration of the Earth's center is not zero, then please tell us the direction & magnitude of the vector.

 

[Dale]

That resistance is called proper acceleration.

No, it is not. Proper acceleration is what is measured by an accelerometer, what you are describing is stress which is measured by a stress transducer.

And, going back to your post 54, yes, if there were no stress at the center of the Earth it would collapse into a black hole.

 

[yuiop]

Nope, you got this backwards:

$$\frac{d\tau_s}{d\tau_c}>1$$

implies:

$$\frac{f_1}{f_2} < 1$$

which is clearly wrong. Remember, clock frequency is the inverse of clock period.

I have consistently referred to $$d\tau$$ as clock rate which is the same as frequency. This is the normal meaning for $$d\tau$$ in relativity.

 

[Passionflower]

No, it is not. Proper acceleration is what is measured by an accelerometer, what you are describing is stress which is measured by a stress transducer.

So you think when you stand with your feet on the Earth's surface there is no proper acceleration but instead stress?
Perhaps you might think a little and perhaps realize that both on a microscopic scale are actually the same thing.

And, going back to your post 54, yes, if there were no stress at the center of the Earth it would collapse into a black hole.

Oh I see, good! So let me ask you this: does stress dilate clocks? Or in different terms: does an object under, as you call it, stress, travel on a geodesic?

Also while we are at it, could you describe a clock or accelerometer that could even in principle be the size of a zero dimensional point?

 

[A.T.]

So you think when you stand with your feet on the Earth's surface there is no proper acceleration but instead stress?

On the surface there is both of them. In the center there is only stress, no proper acceleration.

Also while we are at it, could you describe a clock or accelerometer that could even in principle be the size of a zero dimensional point?

Irrelevant. My example works fine with a normally sized clock & accelerometer floating in a cavity at the center. They will measure zero proper acceleration and a slower clock rate than on the surface, according to General Relativity.

If your gravitational theory predicts a non-zero proper acceleration at the center, then I'm really curious about the direction of the vector.

 

[Passionflower]

a normally sized clock & accelerometer floating in a cavity at the center.

Oh I see, now it is a cavity, you are moving the goalposts!
Did you come to realize that Ricci curvature spoiled your original claim?

If your gravitational theory predicts a non-zero proper acceleration at the center, then I'm really curious about the direction of the vector.

First of all it is not my gravitational theory, I am simply applying the equivalence principle.

For instance consider an atom in the center of the Earth. From all directions there will be a tendency to interfere with the EM forces due to the curvature of spacetime, the resistance to that will result in a proper acceleration in all directions.

 

[Al68]

For instance consider an atom in the center of the Earth. From all directions there will be a tendency to interfere with the EM forces due to the curvature of spacetime, the resistance to that will result in a proper acceleration.

Proper acceleration certainly is the result of net force applied, in this case it's equal to zero, since the net force applied to the '"center of earth" atom is equal to zero.

The individual components of force that result in a net force of zero are irrelevant. Proper acceleration depends only on the net force applied, and is equal to zero in this case.

 

[Passionflower]

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

Oh wait I missed that.

So let me get this right you claim that r=0 in the int. Schwarzschild solution is the center of the ball and there can never be a singularity?

 

[A.T.]

Oh I see, now it is a cavity, you are moving the goalposts!

No, I just gave the third counter example that disproves your wrong statement: "The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling."

While you (as usual) fail to address the counter example.

Did you come to realize that Ricci curvature spoiled your original claim?

No. My first counter example is still valid. GR predicts zero proper acceleration for the center of a solid sphere, because the worldline of this point is a geodesic.

First of all it is not my gravitational theory

It clearly contradicts GR.

a proper acceleration in all directions.

Ohhh... so the non-zero proper acceleration vector is pointing in all directions simultaneously? Fascinating...

So, what is its magnitude then?

 

[Dale]

Perhaps you might think a little and perhaps realize that both on a microscopic scale are actually the same thing.

No, they are not. For one thing they have completely different units. In SI units proper acceleration is in m/s² and stress is in kg/(s²m). Additionally, stress is a tensor field with 9 elements at each event in spacetime, and proper acceleration is a vector with 3 elements at each event along a single worldline. It doesn't matter if you are looking at a microscopic scale or not, they are not the same thing at all.

So let me ask you this: does stress dilate clocks?

Stress is part of the stress-energy tensor which curves spacetime according to the EFE.

Or in different terms: does an object under, as you call it, stress, travel on a geodesic?

Whether or not an object is under stress is irrelevant to whether or not it is traveling on a geodesic. What is relevant is if the sum of the real forces is non-zero.

So let me get this right you claim that r=0 in the int. Schwarzschild solution is the center of the ball and there can never be a singularity?

Yes, that is correct. The singularity at r=0 is for the exterior Schwarzschild solution, not the interior one.